A charged particle $($ charge $=q$; mass $=m)$ is rotating in a circle of radius $R$ with uniform speed $V$. Ratio of its magnetic moment $(\mu)$ to the angular momentum $(L)$ is

• Correct answer
  A

  $\frac{q}{2 m}$
• B

  $\frac{q}{m}$
• C

  $\frac{q}{4 m}$
• D

  $\frac{2 q}{m}$

Solution

The magnetic moment of charge, $\mu=$ niA
and angular momentum of charge, Here, $$ \begin{aligned} L & =m \cdot v \cdot R \\ n & =1 \\ Current, i & =\frac{q}{T} \end{aligned} $$ Time-period, $\quad T=\frac{2 \pi R}{V}$ Equivalent current, $\quad i=\frac{q}{2 \pi R / v}=\frac{q \cdot v}{2 \pi R}$ Area of loop or circle, $A=\pi R^2$ $$ \begin{array}{rlrl} & & \mu & =1 \times \frac{q V}{2 \pi R} \times \pi R^2=\frac{q V R}{2} \\ \Rightarrow & L & =1 \omega=m R^2 \times \frac{V}{R}=m V R \end{array} $$ $I=$ Moment of inertia of loop $$ \therefore \quad \frac{\mu}{L}=\frac{n i A}{m v R}=\frac{q v R / 2}{m v R}=\frac{q}{2 m} $$

Hence, the correct answer is option: 'A'.